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April 5

What's the name for this?

I came up with a curious mathematical structure but can't find anything about it, but I guess it has already been studied by others, and there must be a name for it.

Let $(X,d)$ be a metric space that also has a probability measure $(X,{\mathfrak {B}},\mu )$ . Define a kind of distance on measurable sets as $\delta (A):=\int _{x\in A,y\in A}d(x,y)d\mu$ ., and $\beta (A):=\inf {\big \lbrace }\sum \delta (A_{i})|\bigcup _{i=1}^{N}A_{i}=A{\big \rbrace }$ .

The function $\beta$ is intended to be low when the points of A are strongly clustered.

Whats the name for this structure? 95.113.84.102 (talk) 10:48, 5 April 2014 (UTC)[reply]

Zero? Sławomir Biały (talk) 11:40, 6 April 2014 (UTC)[reply]

Yes for countable sets. For countable sets define

\beta _{N}(A):=\inf {\big \lbrace }\sum \delta (A_{i})|\bigcup _{i=1}^{N}A_{i}=A{\big \rbrace }

with fixed N. 93.132.27.1 (talk) 11:44, 6 April 2014 (UTC)[reply]

I'm not sure what role countability has. For the unit interval with the uniform measure, it's zero. For the real line with a normal distribution, it's zero. For a delta measure, it's zero. I think you would need a very pathological metric space for this to be non-zero. (Here I am assuming that the above "inf" is taken both with respect to coverings of A and the number N of sets in the cover. If you only cover with a fixed number of sets, then I have no idea.) Sławomir Biały (talk) 12:25, 6 April 2014 (UTC)[reply]

OK, so the above definition is probably not what I wanted it to be. What I wanted was some function that tells me how strong a set of points is clustered, and possibly also a way to tell the number of clusters. Standard deviation tells me how strong points are concentrated about the center which is gained by averaging the positions. but in case my points are the superposition of 2 or more normal distributions I can't see that from standard deviation. 93.132.27.1 (talk) 12:29, 6 April 2014 (UTC)[reply]

It probably depends on the specific application. My first idea would be to consider something from electrostatics: think of the probability distribution as charge. The total energy is greater for more concentrated charges. In good cases, this is given as an explicit integral of a potential (which, in turn, is also given as an explicit integral of the charge distribution). But precise formulas might not exist at the specified level of generality though. Sławomir Biały (talk) 23:06, 6 April 2014 (UTC)[reply]

Numbers whose prime factors include only 2 and 3

How we call this ensemble? What's the best algorithm that test if a given number is in that ensemble or not? Thanks for your answers. Hunsu (talk) 11:57, 5 April 2014 (UTC)[reply]

That would be {-6,0,6} . Unless you mean something else? Like numbers whose prime factors include only 2 and 3? Staecker (talk) 12:03, 5 April 2014 (UTC)[reply]

It's all numbers of the form

2^{a}3^{b}

for a,b non-negative integers. The best computer algorithm would probably exploit the binary representation of the number, which presumably is already available on the computer hardware. Truncate off all trailing zeros and if the remaining number is a power of 3, then the number you started with must be that power of three times a power of two (the power of two corresponding to the number of digits that you truncated on the first step). Sławomir Biały (talk) 12:07, 5 April 2014 (UTC)[reply]

I changed my question, it was a mistake. Hunsu (talk) 12:11, 5 April 2014 (UTC)[reply]

It has the snappy name of A003586. See also Smooth number. AndrewWTaylor (talk) 19:39, 5 April 2014 (UTC)[reply]

April 6

sum of reciprocals of numbers with k different prime numbers diverges for all k?

Resolved

Is it true that for all k, the sum of the reciprocals of the numbers which contain k factors all different diverges? So for example for k = 4, the sequence starts out 1/(2*3*5*7) + 1/(2*3*5*11) + 1/(2*3*5*13) + 1/(2*3*7*11) +... Naraht (talk) 14:20, 6 April 2014 (UTC)[reply]

It diverges. Use induction on k. Sławomir Biały (talk) 14:43, 6 April 2014 (UTC)[reply]

Why on k? By the time we get to large 4 factor numbers, we will be adding 1/(2*3*5*553105253) which is quite small. It isn't obvious to me. -- SGBailey (talk) 19:14, 6 April 2014 (UTC)[reply]

Well, so the first thing you have to know is that Σ(1/p) diverges, albeit quite slowly (IIRC the sum of the first n reciprocal primes is asymptotic to a constant times log(log(n)). From there it should be pretty easy. --Trovatore (talk) 19:42, 6 April 2014 (UTC)[reply]

I am not going by the induction argument, but just as a nonrigorous intuition on SGBailey's statement: given that the sum of the reciprocals of the primes diverges, and the sum without the constraint that the k factors be different is the kth power of this, and hence this diverges. Those denominators that have duplicated factors forms a vanishingly small part of the series, so removing them to get the original sum should not change the divergence. While this is hand-wavy, it should supply an adequate countering intuition. —Quondum 19:55, 6 April 2014 (UTC)[reply]

You're making it waaayy too hard. It's trivial; you're missing an obvious point. --Trovatore (talk) 20:17, 6 April 2014 (UTC)[reply]

Which is why you're giving no hints. Yes, I can see a simple proof for all k, but it does not use induction. —Quondum 20:59, 6 April 2014 (UTC)[reply]

Alright, let's go ahead and give the hints, or actually the whole answer, now that I've had my coffee and French toast. The point is that, taking the k=4 case as an example, the series contains a subseries that looks like 1/(2*3*5) times the series (1/7+1/11+1/13+1/17+...). The latter series diverges because it's just the sum of the reciprocal primes, leaving off the first three terms. And if you multiply a divergent sequence by a nonzero constant, it's still divergent. I agree, it doesn't "use induction", at least at the level of natural-language proof (it probably does if you try to formulate it as a formal derivation in Peano arithmetic, but then so does practically everything, so that's not very interesting). --Trovatore (talk) 21:24, 6 April 2014 (UTC)[reply]

Thank you, that makes sense. I was assuming that something inductive was needed, but the subseries as a constant times a subset of the primes minus k-1 terms it much more direct.Naraht (talk) 22:28, 6 April 2014 (UTC)[reply]

Small circles on a 2-sphere

I am not a mathematician. I am involved in some applied aspects due to my hobby so to speak although it is more serious than a simple hobby. Also I want to reassure you, it is not a homework. I read the article "Sphere" in the Wikipedia and could not find what I need.

Imagine a 2-sphere of radius R = 1 with a North Pole and a great circle which is the Equator. Given a point on the sphere which is distinct from the North Pole and the Equator (it is defined by an angle θ between a radius pointing to the point in question and the one going to the North Pole) I want to know the length (in relative units) of a small circle running through this point parallel to the equator. I would appreciate if someone would give me this answer. Thanks, - --AboutFace 22 (talk) 15:01, 6 April 2014 (UTC)[reply]

Well, the length of the equator is 2piR, so just 2pi in your R = 1 case. To reduce that by the angle from the north pole, you'd need to put a sine in there. I'm thinking it would be squared, so sin²(θ)2pi. Can somebody else verify this ? (I believe this works for circles in the southern hemisphere, too.) StuRat (talk) 15:11, 6 April 2014 (UTC)[reply]

No, it's

2\pi R\sin \theta

. Looie496 (talk) 15:14, 6 April 2014 (UTC)[reply]

Yes, I just grabbed a globe and verified that. I measured 29 inches at the equator and 20.5 inches at the 45 degree mark, giving me a ratio of 0.7069, right on with the sin(45) = 0.7071. StuRat (talk) 15:19, 6 April 2014 (UTC)[reply]

Wow! So fast! Thank you very much. StuRat is here as always! I don't see the proof though. It seems to be more like intuitive? I hope you guys are correct. It is definitely correct in two extreme cases, theta = 0 and theta = 90. Looie496, thanks. Thanks again. --AboutFace 22 (talk) 15:24, 6 April 2014 (UTC)[reply]

The proof is trivial, although our latitude article is a bit technical to serve as a good starting point. See instead (for instance) this diagram; if you complete the (right) triangle formed by the radius to a point on the surface, a line through that point parallel to the equatorial plane, and a portion of the axis, you will see that the distance from the surface point to the axis is

R\sin \theta

(your θ is the colatitude that mathematicians typically use). That distance is the radius of the parallel at that latitude, so the circumference is

2\pi R\sin \theta

, as given. --Tardis (talk) 16:29, 6 April 2014 (UTC)[reply]

Oh, and note that this is only for a theoretically perfect sphere. The Earth is actually a lumpy oblate spheroid, so your mileage may vary. StuRat (talk) 17:09, 6 April 2014 (UTC)[reply]

Again, I appreciate it. Thanks, --AboutFace 22 (talk) 19:58, 6 April 2014 (UTC)[reply]

Well, I am back here. Either there is something wrong with the formula or I am making a mistake somewhere. First I have a small globe in front of me. Then I decided to calculate the latitude as you call it or a small circle on the sphere which is just 3 degrees removed from the North Pole. It is a tiny distance, in fact. This is what I get. The sine ( θ ) = 0.052359... I don't know how to use calculators, it is easy for me to write a short piece of code, so it is C#. Then using your formula I multiply 6.28 * 0.052359 and I get 0.3284. It seems too large a value for me. Could it be true? Thanks, --AboutFace 22 (talk) 21:29, 6 April 2014 (UTC)[reply]

Yup, correct (you missed a digit) for the circumference (what I assume you mean by "length") of the circle. This is 5.2% of the circumference of the equator. Colatitude: 3°, Latitude: 87°, circumference of circle of latitude: 0.3288365⋅R, where you took R = 1. —Quondum 21:53, 6 April 2014 (UTC)[reply]

Quondum, thank you but I do not understand what you are trying to say? First you said "correct." In what sense am I correct? That I made a mistake or the result I posted is correct? And what digit did I miss? Could you be more specific? Yes, I am talking about circumference. It is the length of the small circle I meant. On the other hand my result appears to be around 5% of the Equator (great circle). --AboutFace 22 (talk) 22:05, 6 April 2014 (UTC)[reply]

The formula looks right to me. To convince yourself that everything is OK, try what I did: Get a large globe, run a string along it at the equator, and measure the length of the string. Then repeat this while making a circle 3 degrees from the N pole (it helps if your globe has latitude markings). The ratio of the two lengths should be the sine of 3 degrees.

Note that a larger globe and larger angle on the globe from the pole will both reduce the relative error. StuRat (talk) 22:19, 6 April 2014 (UTC)[reply]

The value 0.32884 is the correct value for the circumference of a circle around the north pole at colatitude 3° on a sphere of radius 1. Your value differed from this in that it omitted one of the '8' digits. Perhaps the little circle seems less that 5% of the size of the equator because the area of the discs differs so much (area ratio 1:365). —Quondum 22:22, 6 April 2014 (UTC)[reply]

Thank you StuRat and especially Quondum. Now it is all clear. A rephrased statement did the trick. --AboutFace 22 (talk) 00:41, 7 April 2014 (UTC)[reply]

April 7

How can I integrate these two rational functions?

Help me in integrating these two rational functions (w.r.t. x).
1) 1/(x^3 + 1) and
2) 1/(x^4 + 1) — Preceding unsigned comment added by 117.242.108.241 (talk) 13:11, 7 April 2014 (UTC)[reply]

This looks like homework, but in case it is not, use Partial fraction decomposition and integrate over the parts. 95.112.167.155 (talk) 13:42, 7 April 2014 (UTC)[reply]

Any rational function can be integrated by a combination of one or more of (1) polynomial long division, (2) partial fractions, (3) completing the square. For 1/(x^3+1) factor the denominator as (x+1)(x^2-x+1) and apply partial fractions. For the second, factor the denominator as

(x^{2}+{\sqrt {2}}x+1)(x^{2}-{\sqrt {2}}x+1)

and apply partial fractions. Sławomir Biały (talk) 13:49, 7 April 2014 (UTC)[reply]

Or just write down the sum of all the principal parts of the Laurent expansions around each pole. You have that. 1/(x^n + 1) has singularities at x_k = exp[(2k+1)pi i/n]. We have:

Lim x to x_k of (x-x_k)/(x^n + 1) =1/(n x_k^(n-1)) = 1/n exp[-(2k+1)(n-1)pi i/n]

This is then the coefficient of 1/(x-x_k) in the Laurent expansion about x = x_k. The partial expansion is thus given as:

1/(x^n + 1) = sum from k = 0 to n-1 of 1/n exp[-(2k+1)(n-1)pi i/n] 1/(x-exp[(2k+1)pi i/n])

It's not difficult to integrate this term by term and recast this in a manifestly real form using the relations between logarithms of complex arguments and the arctan function. Count Iblis (talk) 18:57, 7 April 2014 (UTC)[reply]

How to check my antiderivative?

Is there any way(or method or trick) to check whether the antiderivative I got after integrating a function is right or wrong? I know it can be done by differentiating the antiderivative and matching it with the original function, but I am searching for some other method. 117.242.108.241 (talk) 16:07, 7 April 2014 (UTC)[reply]

Well, you could use graphic methods to get an approximation, by looking at the slope of a curve (then curvature) or the area under the curve, depending on which direction you are going. What's good about this method is that it's totally independent of other mathematical methods, so a mistake made there will not be replicated graphically. On the negative side, the graphic method can only check the curve within a certain domain/range, not over it's entire length. StuRat (talk) 16:11, 7 April 2014 (UTC)[reply]

The feat of making it's mean two completely different things in the same post should not go unnoticed. 84.209.89.214 (talk) 17:25, 7 April 2014 (UTC)[reply]

What I sometimes do is numerical differentiation. I did that just yesterday when I wrongly integrated arccos(1/t). I needed the integral to do some calculations, so I had programmed the antiderivative in my programmable calculator. Then it was easy to estimate the derivative in some point as

[F(x+h) - F(x-h)]/(2h)

for small h, which is much more accurate than the estimate [F(x+h) - F(x)]/h, so you can make h larger which leads to less loss of significant terms. I caught a stupid error that way and then corrected the mistake. Examples:

[Log(10.01) - Log(9.99)]/0.02 = 0.1000000333333533

[exp(1.01)-exp(.99)]/0.02 - exp(1) = 0.000045304923665

Count Iblis (talk) 19:08, 7 April 2014 (UTC)[reply]

Some of these rather simple formulas are remarkably precise, as Count Iblis mentioned above. For example, an area below the curve between x=L and x=R can be approximated by (R–L) f((R+L)/2) quite precisely. This is known as the rectangle rule. As long as the slope (the derivative of f) doesn't change sign between L and R, the "missing" area will at least partially cancel the "spurious" area at the other side of the rectangle. In most cases, it is even better than the other simple formula, (R–L) (f(R)+f(L))/2, which takes two f values.

I found that with well-behaved functions, most parts of the curve are approximately parabolic; thus, the true area is usually somewhere between the two rough estimates and closer to the rectangle rule. If the second derivative doesn't change sign, the difference is a rigorous upper bound of the error term. - ¡Ouch! (^{hurt me} / _{more pain}) 09:07, 10 April 2014 (UTC)[reply]

Can't you calculate exactly how long an infinite amount of monkeys would take to type "Hamlet"?

The normal version of the theorem I understand; eventually a lone typing monkey will write Hamlet. What I don't get is why an infinite amount of typing monkeys are said to eventually type Hamlet, when a more accurate statement is as quickly as it is possible to type Hamlet.

If there are an infinite amount of typing monkeys, then a set infinite amount of them will successfully type the first letter of Hamlet on their first try (as they will all letters). A set infinite amount of those monkeys will then type the second letter, and so on until a set of them have typed Shakespeare's Hamlet from the millisecond they sat down and began.

Therefore, the "eventually" is actually a known number. Just calculate how much time it takes to type a single character on a typewriter (T), and the amount of characters in Hamlet (C), then it would take an infinite amount of typing monkeys T*C long to do it. Right? — Preceding unsigned comment added by 50.43.180.176 (talk) 22:20, 7 April 2014 (UTC)[reply]

An infinite "amount" of monkeys? Do you buy your monkeys by the gallon, or what? Actually, if you have an infinite number of these idealized monkeys (really random-number generators), then it's true that almost surely there will be some monkey that types Hamlet straight through without a single false start. However, almost surely is different from surely. They could actually type forever and never type Hamlet, though the probability of that is zero.

Anyway, these calculations are a little beside the point, I think. --Trovatore (talk) 22:42, 7 April 2014 (UTC)[reply]

Well, there are an infinite number of editors editing Wikipedia, with identical copies of each of us about 10^(10^29) meters apart. Count Iblis (talk) 23:41, 7 April 2014 (UTC)[reply]

Except that while they remain identical, they're typing the same thing... —Quondum 00:00, 8 April 2014 (UTC)[reply]

I had an unusually unsuccessful session at the keyboard yesterday. Unfortunately, the result was nothing like Hamlet. YohanN7 (talk) 23:59, 7 April 2014 (UTC)[reply]

But how much are your mistress' eyes like the Sun? --Trovatore (talk) 00:02, 8 April 2014 (UTC) [reply]

Not that this changes the calculations in the original question, but I believe that the saying is that an infinite number of monkeys, typing for an infinite amount of time, will eventually type all of Shakespeare (not just one play, Hamlet). See here: Infinite monkey theorem. Joseph A. Spadaro (talk) 15:50, 8 April 2014 (UTC)[reply]

But, yes. I think that your original calculation (T x C) is correct. If we have an infinite number of monkeys, an infinite number will type Hamlet correctly on the first try; an infinite number will make 1 mistake; an infinite number will make 2 mistakes; ... and so on, until an infinite number will make every mistake possible (and, hence, never type Hamlet at all). But, as one editor above said, this all misses the point of the theorem. The point is not to calculate the amount of time it would literally take; but, rather to show that this feat (as is any other feat) is indeed possible to achieve. Thanks. Joseph A. Spadaro (talk) 15:58, 8 April 2014 (UTC)[reply]

With probability 1, all those things will happen. But not for sure. --Trovatore (talk) 19:24, 8 April 2014 (UTC)[reply]

An important ingredient that nobody's mentioned is that the monkeys need to all be typing "at random" in such a way that after T × C steps they have collectively produced all possible texts of length C. (This is not guaranteed just by saying there's infinitely many of them.) Even so, there are many (infinite length) texts which will never be produced by such a collective because of Cantor's diagonalization argument. Staecker (talk) 16:05, 8 April 2014 (UTC)[reply]

Oh, are you one of those poor souls who only have a countably infinite number of monkeys? PrimeHunter (talk) 16:15, 8 April 2014 (UTC)[reply]

Excuse my Hebrew, but "

\aleph _{0}B

, or not

\aleph _{0}B

", is that the question?" - ¡Ouch! (^{hurt me} / _{more pain}) 08:37, 9 April 2014 (UTC)[reply]

If we start imagining infinite numbers of physical objects then I think things get weird according to quantum theory. You don't need monkeys or typewriters. Just place an infinite number of ink bottles in front of stacks of paper, and an infinite subset of them should transform into Hamlet within a millisecond. Then do away with the ink and paper. PrimeHunter (talk) 16:09, 8 April 2014 (UTC)[reply]

@PrimeHunter: If you want to go down that route, you only need one bottle of ink and one stack of paper in the real world, under the MWI. —Quondum 16:30, 8 April 2014 (UTC)[reply]

You only need a cloud of hydrogen, because even at room temperature, the probability of nuclear fusion is nonzero, so we can get our ink and paper from pure hydrogen. :P That'd explain the pricing of HP cartriges, too.

However, one huge cloud of hydrogen won't do the trick; we'd need several smaller clouds, lest it collapse into one dense mtherfucker and take our copy of Hamlet with it.

Uh-oh, we should stop now. - ¡Ouch! (^{hurt me} / _{more pain}) 06:27, 10 April 2014 (UTC)[reply]

@Staecker: The requirement of an infinite number of typing monkeys having produced every single text of a given length in the shortest possible time is still unity, because the number of texts of a given length is finite (assuming a finite alphabet). The probability of producing all finite texts (of all lengths) in infinite time might be more interesting, as this looks more like the indeterminate form 1^∞. —Quondum 16:30, 8 April 2014 (UTC)[reply]

If you have a countable set of events, each of which has probability 1, then the probability that all the events occur is also 1. This follows from countable additivity. --Trovatore (talk) 01:53, 9 April 2014 (UTC)[reply]

Would you care to show how this applies? Let's assume you only have a countably infinite number of monkeys that start typing in succession one keystroke apart, and you require that every possible finite text occurs for some monkey from the start of its typing. I'd say that finding a threshold length that is a function of time that tends to infinity such that the probability of having all texts up to that length tends to certainty is going to be quite a challenge. —Quondum 02:31, 9 April 2014 (UTC)[reply]

You're making it too hard. Assuming there are only countably many keys on the typewriter, there are only countably many finite texts that can be created. For each such text, the probability that it does not show up is 0. Therefore the probability that at least one of them does not show up is bounded by the sum of an infinite series all of whose terms are 0. --Trovatore (talk) 02:48, 9 April 2014 (UTC)[reply]

I don't agree. You cannot choose the order in which you take two limits. The probability of any given text length T being typed by a given monkey by time T drops very rapidly with time: much faster than 1/T. Even if we start all monkeys typing at the same time, but simply calculate the sequence of probabilities of all texts up to length T having been typed by time T as T tends to infinity, we get a sequence of which the limit is very strongly zero. Yet we've included the probability of all finite texts having been collectively typed by countably infinite monkeys in countably infinite time. —Quondum 04:56, 9 April 2014 (UTC)[reply]

It's not a question of agreeing or not. I gave a proof. Limits have nothing to do with it, and time has nothing to do with it. The probability measure is countably additive, which means the probability of the join of countably many incompatible events is equal to the sum of the probabilities of those events. Drop "mutually exclusive", and it becomes less-than-or-equal-to, but less-than-or-equal-to zero is zero. --Trovatore (talk) 07:37, 9 April 2014 (UTC)[reply]

But what happens if you replace the monkeys with Wikipedia editors? YohanN7 (talk) 12:03, 9 April 2014 (UTC)[reply]

I apologize for my tone in the above. Quondum, can you say what you take it to mean that the probability of an event has a given value, when the condition can neither be verified nor falsified in finite time? I'm using the notion from measure theory; it's not defined in terms of a limit as time goes to infinity, though you might with sufficient cleverness be able to express it in some such way. (Possibly relevant is Fatou's lemma, though I don't actually remember the statement of the lemma, so it might not be.) --Trovatore (talk) 20:55, 9 April 2014 (UTC)[reply]

It's not clear to me that the concept of "the probability that all finite texts will be typed in infinite time" even has a meaning, and I'm not familiar with measure theory or the necessary math to deal with this one, so I can't comment. Let's just say I find the whole idea rather challenging; I'm not going to argue the point. —Quondum 04:06, 10 April 2014 (UTC)[reply]

A better defined question is the following. Since the decimal expansion of almost all real numbers will contain an infinite number of Hamlets, one can ask where the first Hamlet contained within the decimal expansion of pi is located. Count Iblis (talk) 19:33, 8 April 2014 (UTC)[reply]

Abstract Algebra: GCD

I have a simple problem, but I am unsure about one main idea. The question is for any integer a, b, c prove that gcd(a, b)=gcd(a, b+xa) for any x in Z.

My goal is to show that if the gcd(a,b) =s and the gcd((a, b+xa)=s then they are equal to each other. My proof is as follows:

Then, I go on to prove that gcd(a, b+xa)=k by using the definition of gcd.

Can I use the properties of gcd attained from assuming gcd(a,b) =s in the second part of proof (proving gcd(a, b+xa)=s)??

For example, to prove if u|a and u|b+xa then u|k, can I use the fact that the gcd(a, b)=s can be written as a linear combination s=ax+by? Am I going wrong somewhere here? — Preceding unsigned comment added by Abstractminter (talk • contribs) 22:47, 7 April 2014 (UTC)[reply]

April 8

One Last Coincidence

{\begin{cases}\displaystyle ~\int _{0}^{1}{\Big (}1-{\sqrt[{^{\frac {1}{2}}}]{x}}{\Big )}^{^{1-{\frac {1}{2}}}}dx~=~{\frac {\pi }{4}}\\\\\displaystyle ~\int _{0}^{1}{\Big (}1-{\sqrt[{x}]{x}}{\Big )}^{1-x}dx~\approx ~{\sqrt {\frac {\pi }{4}}}~\approx ~\int _{0}^{1}{\Big (}1-{\sqrt[{^{\frac {1}{3}}}]{x}}{\Big )}^{^{\frac {1}{3}}}dx\end{cases}}

As usual, my question is whether there might not be something deeper to these `coincidences`. — 79.113.255.148 (talk) 00:29, 8 April 2014 (UTC)[reply]

Lorentz group: Exponential mapping onto?

Hi!

Is the map $exp: so (3;1) \to SO + (3;1)$ onto?

I know it is for the rotation subgroup. I also know that every LT $Λ$ can be written as $Λ = BR$ where $B$ is a pure boost and $R$ is a pure rotation. Then $Λ = e K e J$ , where $K$ and $J$ are suitable "generators" of boosts and rotations respectively.

Is it generally true then that $Λ = e M$ for some $M \in so (3;1)$ ?

The answer is definitely affirmative near the identity by the Baker-Campbell-Hausdorff formula. $M$ is then a bracket series in $K$ and $J$ (not $K + J$ ).

How about far from the identity? YohanN7 (talk) 11:29, 8 April 2014 (UTC)[reply]

Yes, the exponential map is surjective in this case. Whenever the Lie group is compact, the exponential map is a surjection onto the identity component. I don't know of an easy proof of this, but you can use the Hopf–Rinow theorem to see it. In the case of the Lorentz group

\mathrm {O} (3,1)

, the connected component containing the identity is

\mathrm {SO} ^{+}(3,1)

. --Sam^Talk 12:51, 8 April 2014 (UTC)[reply]

I believe your conclusion may be correct, but the reason you give isn't, because the Lorentz group is not compact. YohanN7 (talk) 13:22, 8 April 2014 (UTC)[reply]

The onto character of

exp

in the case of

SO(3)

and other compact classical groups is easiest seen utilizing the fact that the group elements are all conjugate to matrices of a special form. The representatives of these conjugacy classes are then explicitly seen to be one-parameter subgroups. YohanN7 (talk) 13:35, 8 April 2014 (UTC)[reply]

The same reasoning will apply for the Lorentz group. It's a quotient of the complex semi simple group SL(2,C), and in that group you can use the Jordan decomposition. (It's probably true for all complex semisimple connected groups for basically the same reason, but I haven't really thought about the details). Sławomir Biały (talk) 13:52, 8 April 2014 (UTC)[reply]

The exponential map for

\mathrm {SL} _{2}(\mathbb {C} )

is not surjective, for instance the matrix

q={\begin{pmatrix}-1&1\\0&-1\end{pmatrix}}

is in Jordan normal form and is not in the image of the exponential map. --Sam^Talk 14:12, 8 April 2014 (UTC)[reply]

Oh, of course you're right. My argument only works if you have nonzero trace. Sławomir Biały (talk) 15:19, 8 April 2014 (UTC)[reply]

Oops, sorry about that YohanN7! I was too hasty. Yes, another proof for compact groups is to use that every element is contained in a Cartan subgroup, and that these are all conjugate. This is still not helpful in this instance though.

Here's an argument. The exponential map

\exp \colon {\mathfrak {sl}}_{2}(\mathbb {C} )\to \mathrm {SL} _{2}(\mathbb {C} )

is nearly surjective:

\exp \colon {\mathfrak {sl}}_{2}(\mathbb {C} )\to \mathrm {PGL} _{2}(\mathbb {C} )=\mathrm {SL} _{2}(\mathbb {C} )/\{\pm \mathrm {I} \}

is surjective. From this the result follows for

\mathrm {SO} ^{+}(3,1)\cong \mathrm {PGL} _{2}(\mathbb {C} )

because of the exceptional isomorphism

\mathrm {SL} _{2}(\mathbb {C} )\cong \mathrm {Spin} (3,1)

. I don't know how you would see it in more generality. --Sam^Talk 14:10, 8 April 2014 (UTC)[reply]

My purpose is to come with a reliable statement regarding this in Representation theory of the Lorentz group. Either of a "reliable source" or a simple proof would do. I haven't seen the statement (or its negation) being made anywhere in the literature.

Would you be able to come up with something less general without explicit or implicit isomorphisms? The exponential mapping is (in this context), after all, a map from a Lie algebra to the connected group generated by it, not to another isomorphic group or some quotient. Consider, for reference, the proof below for compact classical groups:

Consider a compact classical group

K

. Let

g \in K

be arbitrary. Then

g = hch -1

where

h \in K

and

c = e X

with

X \in k

, the Lie algebra of

K

(conjugation properties of compact classical groups). Now, by following the definition of

exp

,

h e X h -1 = e hXh -1

and

hXh -1 \in k

by using properties of the adjoint representation of

K

.

A proof like this, I could shorten/translate to words and give ample references to the literature/internal links for each step. (Unfortunately, it doesn't apply.) Proofs using hidden isomorphisms aren't convincing. YohanN7 (talk) 14:58, 8 April 2014 (UTC)[reply]

Let

π: SL(2, C) \to PGL(2, C)

be the quotient map. Could I then easily prove that for each

p \in PGL(2, C)

, there is at least one (exactly one would be nice) element of the fiber

π -1 (p)

that is in the image of

exp: sl (2, C) \to SL(2, C)

? Your answers above suggests that this is the case, except perhaps for "ease" of proof. That would solve the problem (using some isomorphisms). YohanN7 (talk) 16:47, 8 April 2014 (UTC)[reply]

I'm not sure what you're asking. Once you know that the exponential map

\exp \colon {\mathfrak {sl}}_{2}(\mathbb {C} )\to \mathrm {PGL} _{2}(\mathbb {C} )

is surjective, the rest follows. That it is surjective for

\mathrm {PGL} _{n}(\mathbb {C} )

follows from the fact that it is so for

\mathrm {GL} _{n}(\mathbb {C} )

, by the Jordan normal form considerations mentioned above.

There can be either 1 or 2 elements in the image of

\exp \colon {\mathfrak {sl}}_{2}(\mathbb {C} )\to \mathrm {SL} _{2}(\mathbb {C} )

in a fiber, e.g.

\{\pm \mathrm {I} \}

. --Sam^Talk 17:46, 8 April 2014 (UTC)[reply]

I don't know how to prove the more general result for

\mathrm {SO} (n,1)

, so I don't know how to avoid using accidental isomorphisms. --Sam^Talk 17:48, 8 April 2014 (UTC)[reply]

Thanks, with the

GL(n, C)

-argument, there are no loose ends, but that was new. At any rate, I have found an argument using

SL(2, C)

and its conjugacy classes only (that I can source, Rossman). If I can prove that

p={\begin{pmatrix}1&-1\\0&1\end{pmatrix}}

is in the image of

exp

, then I'm done because

qp -1 = -I

so

p ~ q

under

π

. (Note that I've taken the liberty of naming your matrix of above.) The rest is exactly as in the compact case. YohanN7 (talk) 18:47, 8 April 2014 (UTC)[reply]

And it looks like

p=exp{\begin{pmatrix}0&-1\\0&0\end{pmatrix}}.

YohanN7 (talk) 18:57, 8 April 2014 (UTC)[reply]

Just to explain my "problem" with (hidden) isomormhisms. With these, for a full proof, one needs to display (at least imagine) a commutative diagram, at the worst one has to prove that it commutes. In the present case, with notation as above, there is a theorem. If

π: G \to H

is a homomorphism, then its differential

φ: g \to h

is a homomorphism and

π(exp(X) = exp(φ(X))

. The lhs as a whole is surjective. Therefore the exp on the rhs (not the same exp as on the lhs) is surjective. Since the conjugacy class represented by

q

above was the only problematic class, I'm now completely satisfied. YohanN7 (talk) 20:25, 8 April 2014 (UTC)[reply]

Expressing 2.5cm in mm?

How will you do that? Please explain me, step by step how to do it (everywhere I ask, I get too much of a short answer which doesn't help me understand the way needed for this simple calculation). thank you. 79.177.27.113 (talk) 18:38, 8 April 2014 (UTC)[reply]

Converting between metric units is designed to be "easy," but it can be tedious to explain every little step. Metric conversion doesn't exist as an article, and conversion of units thinks that converting between metric prefixes is too east to mention! So, here's an explanation, hope it helps.

First, you should be familiar with the meter, whose symbol is 'm', and the metric prefix system, which explains how powers of ten are communicated with words like "centi-" or "deci-" and so on. So, 2.5 cm is 2.5 centi meters and centi- means "one one-hundredth", which is (2.5 meters)/100. You could also think of it as (2.5 meters) * 0.01, which is the same thing written a little differently. Now, if you want it it milli meters, you look and see that "milli-" means a factor of 1/1000, or 0.001. Now, we can convert directly to millimeters by just "moving the decimal point" - but that can be tricky for unfamiliar prefixes. So the safest way is to convert cm to m, then m to mm. In this case,

2.5cm=2.5m/100=0.025m

. Now 1 m = 1000 mm, so we can multiply a unit in meters by 1000 to get a unit in mm. So, 0.025 m = 0.025 X 1000 mm = 25 mm.

Finally, we have

2.5cm=2.5m/100=0.025m=25mm

If that explanation doesn't make sense, here's a few other tutorials on the topic [1] [2]. SemanticMantis (talk) 19:07, 8 April 2014 (UTC)[reply]

If you know there are 10 mm in a cm, you can use the unit multiplication method:

_{2.5 cm x} 10 mm _= ?
         cm

The cm's cancel out:

_{2.5 cm x} 10 mm _{= 25 mm}
         cm

StuRat (talk) 23:20, 9 April 2014 (UTC)[reply]

What's happened to my earlier reply on this topic? And is other stuff missing too? HiLo48 (talk) 23:48, 9 April 2014 (UTC)[reply]

I don't see it listed in your contribution history. So, unless there's a bug or an Admin redacted it, that means you weren't able to save it. StuRat (talk) 23:58, 9 April 2014 (UTC) [reply]

Here it is:

"Year 7 students (12 to 13 years old) in Australia are simply told "To convert centimetres to millimetres, just multiply by 10." HiLo48 (talk) 01:29, 9 April 2014 (UTC)"[reply]

HiLo48 (talk) 01:17, 10 April 2014 (UTC)[reply]

April 9

Calculating flow

There is a channel of some incompressible liquid that stretches away indefinitely from the origin (x = 0) in either direction. This is at a constant depth d. A flow is induced (I am not concerned about how) such that the speed of flow at coordinate x is given by v = kx for some positive constant k. Positive v means flow in the direction of increasing x; negative v means flow in the direction of decreasing x. When I do the calculations, I get that after time t the water is at constant depth d/exp(kt), but I can't figure out how the depth at x = 0 can have changed when there is no flow there. I find it quite hard to visualise what ought to happen around x = 0, actually. Am I doing this right? 86.179.5.189 (talk) 20:19, 9 April 2014 (UTC)[reply]

Let the depth be called y. Choose units of time and length such that k = dv/dx = 1 and y₀ = 1. Use letter d for differentiation. The rate of change of depths , dy/dt = −y dv/dx = −y , doesn't depend on v but only on dv/dx which is unity. The differential equation dy/y = −dt is integrated to log(y) = −t+const. Choose the zeropoint of time such that the constant is zero. Then the depth is y = e^−t , independently of x. Bo Jacoby (talk) 03:53, 10 April 2014 (UTC).[reply]

To help visualization, imagine a bucket of water, from which you suddenly remove the sides. The water flows equally in all directions, but the water at the centre has zero radial velocity, only a (depth-dependent) downwards velocity. Your scenario is qualitatively the same. —Quondum 04:22, 10 April 2014 (UTC)[reply]

Here reality differs from theory. In reality you would get all kinds of flow at the center. If you think about it as the molecules of water, there's no way the top molecules can go lower, and remain at the center, without the molecules underneath it moving out of the way. StuRat (talk) 04:34, 10 April 2014 (UTC)[reply]

I don't understand Bo Jacoby's approach. I want an answer that involves d and k. Is d/exp(kt) correct? 86.160.87.195 (talk) 11:05, 10 April 2014 (UTC)[reply]

Odds and evens

I'm reviewing a wagering-proposition where I suspect the house has made a mistake in the odds it is offering. In a twenty-horse race (assuming all of the horses are of equal ability) the result of the first three finishers all being odd- or even-numbered was offered at three-to-one odds. This seems to defy logic, as two of the horses will have to be either even or odd. The proposition seems to boil down to: "Will the third horse (of the remaining eighteen) be the same type of number as the other two?". The true odds of this happening seem to only be slightly less than even, certainly nowhere near three-to-one. I want to rush to put the maximum-bet of ~~$300~~ ($100, my mistake) down on this proposition, but being far from the smartest fellow on the planet, I figured I should double-check this with our learned and perspicacious mathematicians first. Thanks in advance! Joefromrandb (talk) 04:39, 10 April 2014 (UTC)[reply]

It is a little more complicated than that and it is just about 4:1, according to my calculations. For one thing, we are assuming that half of the horses have odd numbers. Also, it depends on the number of horses in the race. For instance if there were only six horses, it would have to be a particular three out of those six. And whatever the odd/even of the first horse is, the second one is a little less than 50% (9/19 with 20 horses). And if those two match, the probability that the third matches the first two is a little less than 50% (8/18 with 20 horses). Bubba73 ^{You talkin' to me?} 04:53, 10 April 2014 (UTC)[reply]

Thanks a lot! Your solution was the way I was originally thinking, before I tripped myself up figuring there was some kind of Monty Hall-paradox going on. With your solution, the true odds seem to be slightly higher than 4–1, giving the house quite a substantial edge by offering the 3–1 odds. Joefromrandb (talk) 05:20, 10 April 2014 (UTC)[reply]

Make the following experiment of thought: Let there be infinitely many horses in the race. (Hey, we just had a huge thread with infinitely many monkeys above, why not horses?) Then there are eight equally possible outcomes; (odd, odd, odd), (odd, odd, even), ..., (even, even, odd), (even, even, even). Only two of these are in your favor. YohanN7 (talk) 05:35, 10 April 2014 (UTC)[reply]

Yes, and it approaches that as the number of horses increases. Bubba73 ^{You talkin' to me?} 05:41, 10 April 2014 (UTC)[reply]

Saltire question

I'm making a flag, and I want to make sure I have the proportions right. Let's imagine that this flag is 1000 units wide and 500 units tall. Each arm of the cross has a thickness (A) of 100 units, and is centered on a line running, at a 26.57° angle, from one corner of the flag to the other. What is B, the length of white along the bottom edge on the left? --Lazar Taxon (talk) 11:40, 10 April 2014 (UTC)[reply]

That would be 50√5, or about 111.80. 86.160.87.195 (talk) 12:46, 10 April 2014 (UTC)[reply]